Conditional probability question - Hold 'em

[pilamsolo]

I have a question regarding conditional probabilities for Texas Hold 'em. Take the following situation:

It is a 10-handed game before the flop and 5 handed on the flop. If I have one ace, and there is an ace on the flop, what is the probability that there is another player in the hand that also holds an ace?

I have tried to calculate using Bayes Theorem, but I'm doing something wrong, so any help is appreciated.

[Nottom]

It depends if your opponents would play any ace or just good ones. Without knowing what hands your opponenets would play this is an impossible question to answer.

[ramjam]

If this is a loose low-limit game where any ace is played but nobody ever raises (even with AA or AK), you can solve the problem analytically.

There are 47 unknown cards, your opponents were dealt 18 of them and there are two aces unaccounted for.

The probability that those 18 cards contain no ace is 28*29/(47*46)= 37.6% (I think!). Or put another way, there's a 62.4% chance that someone else does have an ace.

As nottom says, if your opponents are more selective in what they play, you can't solve the problem without knowing exactly what their calling/raising standards are in every situation - and even then, it would probably be too complex to solve this millennium.

[BruceZ]

The probability that those 18 cards contain no ace is 28*29/(47*46)= 37.6% (I think!).

Correct. That's C(45,18)/C(47,18) = 29*28/(47*46). He asked about 4 opponents, and that would be
C(45,8)/C(47,8) = 39*38/(47*46) = 68.5% that they do NOT have an A. You don't need to use Bayes' theorem.

[BruceZ]

You don't need to use Bayes' theorem.

That is, if you assume random hands. If you are taking into account the probability that the hands that were folded on the flop did not contain an A, then you could use Bayes' theorem, but you would need more specific information about the hands your opponents play. The probability that an A is out would then be much more likely.